Here, the first part of the total color matrix requires only \(\frac{n}{2}+2\)  colors by using the result of Campos and de Mello [1]}. From Lemma 1.1, the graph induced by the remaining edges form a Cayley graph and it is 1-factoriziable. Therefore we can color the remaining edges with \(\Delta (G)-\frac{n}{2}\)  colors. This gives us the full total coloring of the original circulant graph using \(\Delta (G)-\frac{n}{2} + \frac{n}{2}+2=\Delta (G) + 2\)  colors thus satisfying TCC.
